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2 years ago

Which of the following is NOT determined when considering the environment factor of developing the approach?

Business

2 answers:

Aleonysh [2.5K]2 years ago

8 0

Answer:

Financial environment

Explanation:

The environment factors that might affect the company are analyzed in order to proceed to set up a strategy for the firm. This means that all "influencing factors of the environment" are considerd. However, there is no existance of financial environment.

juin [17]2 years ago

6 0

Influencing factors of the environment is not determined when considering the environment factor of developing the approach.

<u> Explanation: </u>

Both in the United States and around the world, business choices are among the prevailing impacts forming natural conditions: what materials, vitality, and life forms will be extricated from the earth, in what amounts and where, how they will be moved and circulated, how scenes and environments will be changed in doing as such, and what will be done to limit and alleviate the effects.

Business choices additionally impact customer decisions, direct an enormous portion of natural research, and decide a significant part of the improvement and dissemination of mechanical advancements.

The aggregate impact of organizations' choices makes numerous responsibilities that are troublesome if not difficult to switch temporarily. Buyer inclinations impact these choices at times, yet just to the degree that they firmly influence the capacity to produce benefit.

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When solving for a present value, all cash flows should be discounted from the A : future to the present. B : the past to the fu

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B the past to the future

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3 years ago

Each machine must be run by one of 19 cross-trained workers who are each available 35 hours per week. The plant has 10 type 1 ma

Mrac [35]

Answer:

The Linear programming model is given as below

Profit Function: $P=90X+120Y+150Z$

Constraints:

$2X+2Y+Z\leq 400$

$3X+4Y+6Z\leq 240$

$4X+6Y+5Z\leq 320$

$\dfrac{2X+2Y+Z}{40}\leq 10$

$\dfrac{3X+4Y+6Z}{40}\leq 6$

$\dfrac{4X+6Y+5Z}{40}\leq 8$

$\dfrac{2X+2Y+Z}{35}+\dfrac{3X+4Y+6Z}{35}+\dfrac{4X+6Y+5Z}{35}\leq 19$

Explanation:

As the question is not complete, the complete question is found online and is attached herewith.

Let the number of product 1 to be produced is X, that of product 2 is Y and product 3 is Z

so the maximizing function is the profit function which is given as

$P=90X+120Y+150Z$

Now as the number of hours in a week are 40 and there are a total of 10 type 1 machines so the total number of machine 1 hours are 40*10=400 hours

As from the given table product 1 uses 2 machine hours of machine 1, product 2 uses 2 machine hours of machine 1 and product 3 uses 1 hour of machine 1 so

$2X+2Y+Z\leq 400$

Now as the number of hours in a week are 40 and there are a total of 6 type 2 machines so the total number of machine 2 hours are 40*6=240 hours

As from the given table product 1 uses 3 machine hours of machine 2, product 2 uses 4 machine hours of machine 2 and product 3 uses 6 hour of machine 2 so

$3X+4Y+6Z\leq 240$

Now as the number of hours in a week are 40 and there are a total of 8 type 3 machines so the total number of machine 3 hours are 40*8=320 hours

As from the given table product 1 uses 4 machine hours of machine 3, product 2 uses 6 machine hours of machine 3 and product 3 uses 5 hour of machine 3 so

$4X+6Y+5Z\leq 320$

Now as the machine 1 is used as 2X+2Y+Z in a week and the week is of 40 hours so the number of machines to be used are given as

$\dfrac{2X+2Y+Z}{40}\leq 10$

Now as the machine 2 is used as 3X+4Y+6Z in a week and the week is of 40 hours so the number of machines to be used are given as

$\dfrac{3X+4Y+6Z}{40}\leq 6$

Now as the machine 3 is used as 4X+6Y+5Z in a week and the week is of 40 hours so the number of machines to be used are given as

$\dfrac{4X+6Y+5Z}{40}\leq 8$

Now the workers are available for 35 hours so the worker available at the machine 1 is given as

$\dfrac{2X+2Y+Z}{35}$

That of machine 2 is given as

$\dfrac{3X+4Y+6Z}{35}$

That of machine 3 is given as

$\dfrac{4X+6Y+5Z}{35}$

As the total number of workers is 19 so the constraint is given as

$\dfrac{2X+2Y+Z}{35}+\dfrac{3X+4Y+6Z}{35}+\dfrac{4X+6Y+5Z}{35}\leq 19$

So the Linear programming model is given as below

Profit Function: $P=90X+120Y+150Z$

Constraints:

$2X+2Y+Z\leq 400$

$3X+4Y+6Z\leq 240$

$4X+6Y+5Z\leq 320$

$\dfrac{2X+2Y+Z}{40}\leq 10$

$\dfrac{3X+4Y+6Z}{40}\leq 6$

$\dfrac{4X+6Y+5Z}{40}\leq 8$

$\dfrac{2X+2Y+Z}{35}+\dfrac{3X+4Y+6Z}{35}+\dfrac{4X+6Y+5Z}{35}\leq 19$

4 0

3 years ago

If the price charged for a candy bar is p cents, then x thousand candy bars will be sold in a certain city, where p = 8 - . how

coldgirl [10]

992 candy bars must be sold to maximize revenue.

<h3>What is revenue?</h3>

The total amount of income generated by the sale of goods and services related to the primary operations of the business is referred to as revenue in accounting.
Commercial revenue is also known as sales or turnover.
Some businesses make money by charging interest, royalties, or other fees.

To find how many candy bars must be sold to maximize revenue:

The price of a candy bar is determined by the quantity sold:

p(x) = 124 - (x/16) where x is in 1000s.

If the candy bar's price is p(x), the revenue function is:

R(x) = p(x) · x = 124 · x - x²/16

Find the solution of R'(x) = 0 to maximize R(x):

R'(x) = 124 - x/8
124 - x/8 = 0
x = 992

Therefore, 992 candy bars must be sold to maximize revenue.

Know more about revenue here:

brainly.com/question/25623677

#SPJ4

The correct question is given below:
If the price of a candy bar is p(x) cents then x thousand candy bars are sold. The price p(x) = 124-(x/16). How many candy bars must be sold to maximize revenue?

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